global smooth solutions to euler equations for a perfect gas · global smooth solutions to euler...

Global Smooth Solutions to Euler Equations

for a Perfect Gas

Magali Grassin

Abstract. We consider Euler equations for a perfect gas in Rd,where d ≥ 1. We state that global smooth solutions exist underthe hypotheses (H1)-(H3) on the initial data. We choose a smallsmooth initial density, and a smooth enough initial velocity whichforces particles to spread out. We also show a result of global in timeuniqueness for these global solutions.

Introduction. We consider Euler equations for a perfect gas:∂tρ+ div(ρu) = 0,

ρ(∂tu+ (u · ∇)u) +∇p = 0,

∂tS + u · ∇S = 0,

(1)

where t ∈ R+, x ∈ Rd and u : Rd × R+ → Rd stands for the velocity, ρ :Rd×R+ → R+ for the density, p = (γ−1)ρe for the pressure, with e the internalenergy of the gas and S : Rd×R+ → R+ for the entropy. The adiabatic constantof the gas is denoted by γ > 1 and d ≥ 1 is the dimension of the space.

We are interested in the existence of global smooth solutions to the Cauchyproblem for (1) with (ρ0, u0, S0) as initial data. There exist few results concerningthis problem, especially when d is strictly larger than one. The choice of initialdata is decisive for this problem and it depends on whether one wants to proveor to disprove global existence. We aim at finding conditions on (ρ0, u0, S0) asweak as possible which ensure the existence of a global smooth solution. In [4],T. Sideris has shown a result of non global existence: the initial density is close toa constant at infinity—the constant should be different from 0—and some globalquantities have to be large. For d = 1, in the isentropic case, we have a 2 × 2system. In this case, some results can be proved using P. D. Lax’s works [3]. Inthe same case with less restrictive conditions, J. Y. Chemin [2] has also proveda result of non global existence: the initial velocity has to be smaller than theinitial sound speed in each point—this quantity depends mostly on ρ0—. In [1],

1397

1398 M. Grassin

D. Serre has proved one result of existence in the multi-dimensionnal case, withγ ≤ 1 + 2/d—which is not a restriction in the realistic case—. One restrictionto this result is that the initial velocity must be close to a linear field. We givehere a global existence result without this particular hypothesis. Note that thecase ρ = 0 does occur, that is to say that there can be vacuum in some area.This requires some attention as we will notice later.

This article is divided in two main parts. The first one deals with the isen-tropic case. This is the simplest one and gives the idea of the methods we use.We give a result of global existence and some estimates on the solution. Weshow also a result of uniqueness for all time, under some regularity assumptionson the solution. We prove also a corollary of the existence result which improvesslightly the hypotheses.

The second part generalizes the previous results to a non-isentropic fluid.The vector of all the spatial derivatives of order k is denoted by Dk, and ∂k

is one of the component of Dk. The differential with respect to x of u(·, t) willbe denoted by Du. We denote by |.|p the norm of Lp(Rd) where 1 ≤ p ≤ ∞, by‖.‖0 the one of L2(Rd) and by ‖.‖m the one of Hm(Rd). We also set ‖.‖X thenorm of the space X = {z : Rd × R+ → Rd | Dz ∈ L∞(Rd), D2z ∈ Hm−1(Rd)}.The transpose of a vector V is denoted by V T .

1. The isentropic case. We first consider Euler equations for an isentropicperfect gas: {

∂tρ+ div(ρu) = 0,

ρ(∂tu+ (u · ∇)u) +∇p = 0,(2)

with p = (γ − 1)ργ and the initial data:{u(x, 0) = u0(x),

ρ(x, 0) = ρ0(x).(3)

In this part, we prove a result of global existence for this problem. Then wegive a corollary of this result. Finally, we solve the problem of uniqueness. Wenow find some conditions on the initial data which ensure the global existence.We use a result of local existence of smooth solutions and some energy estimates.Nevertheless, we need to introduce an approximate problem to have a guess ofthe behaviour of the velocity in our problem. Then, we compare this approximatesolution and our local solution in order to obtain accurate energy estimates. Theapproximate problem is the following:{

∂tu+ (u · ∇)u = 0 on Rd × R+,

u(x, 0) = u0(x) on Rd.(4)

It is obtained by neglecting ρ in (2). The hypotheses we make on the initialdata ensure that there exists a global solution to this problem, and that thisapproximate solution stays close to the solution of (2)-(3).

Global Smooth Solutions to Euler Equations for a Perfect Gas 1399

1.1. Main result. We state the following result:

Theorem 1. Let m > 1 + d/2 and assume that

(H1) ρ(γ−1)/20 is small enough in Hm(Rd),

(H2) D2u0 ∈ Hm−1(Rd) and Du0 ∈ L∞(Rd), i.e., u0 ∈ X,(H3) There exists δ > 0 such that for all x ∈ Rd, dist(Sp(Du0(x)),R−) ≥ δ,(H4) ρ0 has a compact support,

and let u be the global solution of (4). Then there exists a global smooth solutionto (2)-(3), i.e., (ρ, u) such that

(ρ(γ−1)/2, u− u) ∈ Cj([0,∞[;Hm−j(Rd)) for j ∈ {0, 1}.

In (H3), dist stands for the distance and Sp for the spectrum. This meansthat the spectrum of Du0 is uniformly bounded away from the real negativenumbers. To explain our hypotheses, we first make this simple remark: sinceour aim is to found a smooth global solution, we choose a small initial densityand an initial velocity which make particles to spread out. Thus, the choice of u0

gives us the existence of a global solution to the simplified problem (4) close to(2). The main hypothesis on ρ0 is its smallness. The exponent (γ − 1)/2 comesfrom the proof of local existence which is the first step in the proof of Theorem1. It is introduced by the symmetrisation of the system. (H1) is not equivalentto ρ0 ∈ Hm(Rd) but these two assumptions can be linked according to the valueof γ. Remark that we accept that ρ0 vanishes in some area. We get rid of (H4)in the corollary following the uniqueness result.

Note that, although the hypotheses have no direct physical interest, sincemost of the physical problems are given in bounded domains, this result em-phasizes the importance of dispersion in the search of a global smooth solution.Moreover, one strongly expects that a physical solution asymptotically behaveslike the one obtained here. Our last motivation in studying the Cauchy problemrather than the initial boundary value problem is that the mathematical theoryof both is rather poor so far, so that it is legitimate to begin with the simplestone.

The proof is based on local existence of a smooth solution to (2) and energyestimates. But the classical method does not work here. Thus, we consider asimplified problem which gives us a global solution u thanks to our hypothesis(H2)-(H3). Then, we prove that there exists a local solution of our problemfor (ρ0, u0), satisfying (H1)-(H4). We compare the two problems and we useproperties of the simpler one to improve the classical energy estimates. Ourproof is split in three steps. First, we introduce u the global smooth solution toan approximate problem concerning only the velocity. We use a local existencetheorem—cf Chemin [2]—and a local uniqueness property of our problem toobtain a local solution with (ρ0, u0) as initial data. Then we find some estimateson u to precise its behaviour. To conclude, we obtain accurate energy estimateson a certain spatial norm of the difference between the local solution and theapproximate one. Thus we obtain the following estimates on our global solution:

1400 M. Grassin

Theorem 2.Under the hypotheses (H1)-(H4), the solution of Theorem 1satisfies:

• ‖DkU(t)‖0 ≤ K(1 + t)−(k+r), for all 1 ≤ k ≤ m, for all t ≥ 0,• |U |∞(t) ≤ K(1 + t)1−a, for all t ≥ 0,• |DU |∞(t) ≤ K(1 + t)−a, for all t ≥ 0,

where U = (π, u− u)T , K depends on δ, ‖u0‖X , ‖ρ(γ−1)/20 ‖m,

r =

1−

d

2if γ ≥ γc = 1 +

2

d,

γ − 1

2d−

d

2if 1 < γ < γc,

and a = 1 + r + d/2 > 1.

Note that, according to the values of k and d, −(k + r) can be positive.

1.1.1. Local existence and approximate problem. We need to supposethat the density is small to expect a global smooth solution. By neglecting ρand ∇(ργ−1), we obtain the approximate problem:{

∂tu+ (u · ∇)u = 0 on Rd × R+,

u(x, 0) = u0(x) on Rd.

In fact, we have neglected ρ(γ−1)/2 in Hm, therefore in C1. Thanks to (H2)-(H3),there is a global solution u in Cj([0,∞[;Hm−j(Rd)) for j ∈ {0, 1}, defined by:

u(X(x0, t), t) = u0(x0), with X(x0, t) = x0 + tu0(x0).

Note that this problem does not take in account any forces. That is why thechoice of u0 is decisive in the global existence of the solution u. We remark thatDu ∈ L∞(Rd × R+), as we show in the next section.

We want now to construct a local solution to (2)-(3) such that the differencebetween this solution and (0, u) is in C0(Hm(Rd)) ∩ C1(Hm−1(Rd)). Note firstthat, since u0 /∈ Hm(Rd), we can not use directly a general local existencetheorem to obtain a local solution to our problem.

The first step in the proof of local existence consists in the symmetrisationof the system. The symmetrisation must be cautiously chosen because the caseρ = 0 can occur. Following T. Makino, S. Ukai, S. Kawashima [6], we take

π =

√(γ − 1)

4γρ(γ−1)/2

to obtain: {(∂t + u · ∇)π + C1πdiv(u) = 0,

(∂t + (u · ∇))u+ C1π∇π = 0,(5)


where C1 = (γ − 1)/2.Actually, this system is not equivalent to (2) because of the case ρ = 0, but

we can pass from (5) to (2) by multiplying by ρ. Thus if we find a global smoothsolution to this problem, we obtain one solution for (2) such that ρ(γ−1)/2 issmooth. But we loose the property of uniqueness of the solution.

We write (5) in the following way:

∂tV +d∑

α=1

Aα(V )∂αV = 0,(6)

where V = (ρ, u)T , Aα(V ) ∈Md+1(R) is symmetric:

Aα(V ) =

uα 0 . . . C1π . . . 00 uα 0... 0

. . ....

C1π...

. . ....

......

. . . 00 0 . . . . . . 0 uα

.

We now construct a local solution to this problem with initial data satisfying(H1)-(H4).

Let R > 0 such that suppρ0 ⊂ B(0, R). Let ϕ ∈ C∞c (Rd) such that ϕ ≡ 1on B(0, R + 2η), where η is some positive constant. We consider (π0, u0ϕ) asan initial data for the problem (5) and we use the theorem of local existenceof solution for symmetric hyperbolic systems, since (π0, u0ϕ) ∈ Hm(Rd)—seein [2]—. Therefore we obtain (πϕ, uϕ) a solution in Cj([0, Tex[;Hm−j(Rd)) forj ∈ {0, 1}. Note that (0, u) is a solution to (5) with (0, u0) as initial data.

Let

K = {(x, t) | 0 ≤ t ≤ T, x ∈ B(0, R+ η +Mt)}, with:

M = sup0≤t≤Tex−ε

(C1|πϕ|L∞ + |uϕ|L∞), and

T = min(Tex − ε, η/(2M)− ε) for ε > 0 given.

Now we take

(π, u) =

{(πϕ, uϕ) in K,

(0, u) outside K.

We have to show that (π, u) is actually a solution on Rd × [0, T ] of (5) with(π0, u0) as initial data. In fact, (π, u) is a solution in K and outside K. Thus wehave just to show that it is continuous across ∂K. In order to do this, we showthat (πϕ, uϕ) and (0, u) are equal on

D = {(x, t) | 0 ≤ t ≤ T, x ∈ B(x0, η −Mt) for x0 ∈ S(0, R+ η)}.

1402 M. Grassin

R R + η

R + 2η

∂K

(0, u) = (πϕ, uϕ)

(πϕ, uϕ)T

K

(0, u)

(π0, u0) = (0, u0)(π0, u0)

Figure 1: Local existence

This is true thanks to a property of local uniqueness of the solutions of thesystem (5).

Proposition 1. Let V 10 , V 2

0 be two initial data for (5). Assume that V 10 ∈

Hm(Rd). Let V 1 = (π, u)T , V 2 be two associated solutions of (5) defined for 0 ≤t ≤ T0, and let M ≥ sup(x,t)∈Q{(C1|π|+ |u|)(x, t)}, where Q = B(x0, η)× [0, T0].Suppose that V 1

0 = V 20 on B0 = B(x0, η) and take

CT = {(x, t) | 0 ≤ t ≤ T, x ∈ Bt = B(x0, η−Mt)}

for 0 ≤ T ≤ T1 = min(T0, η/M). Suppose moreover that |DV 2|∞ < ∞. ThenV 1 = V 2 on CT1

.

The proof of this proposition is classical. We use the properties of the symmetricsystem (5).

We apply this proposition to V 10 = (π0, u0ϕ)T , which is in Hm(Rd), and

V 20 = (0, u0)T . Then we have V 1 = (πϕ, uϕ)T and V 2 = (0, u)T . We know

that Du ∈ L∞(Rd ×R+), and we have chosen M to satisfy the condition in theproposition. As (π0, u0ϕ) = (0, u0) on each B0 = B(x0, η) for x0 ∈ S(0, R+ η),we have (πϕ, uϕ) = (0, u) on D. This implies that (π, u) is smooth across ∂Ksince D contains ∂K.


Thus we have found (π, u) a local solution to our problem such that

(π, u− u) ∈ Cj([0, T [, Hm−j) for j = 0, 1.

1.1.2. Estimates for the approximate solution.We now precise ourknowledge of u. We use hypothesis (H3), which requires some uniformity, to ob-tain estimates on the spatial norm of the derivatives of u. We have the followingresult:

Proposition 2. Suppose (H2), (H3). Let u be the global smooth solutionof (4). Then:

(i) Du(x, t) =1

(1 + t)I +

1

(1 + t)2K(x, t), for all x ∈ Rd, all t ∈ R+,

(ii)∥∥D`u(·, t)

∥∥0≤ K`(1 + t)d/2−(`+1), for 2 ≤ ` ≤ m+ 1,

(iii)∣∣D2u(·, t)

∣∣∞≤ C(1 + t)−3,

with I = IdRd and K : Rd × R+ → Md(R), |K|L∞(Rd×R+) ≤ M , where M , C,and K`, for 2 ≤ ` ≤ m + 1, are some positive constants which depend on m, d,δ, ‖u0‖X .

We remark that the decay in t of the derivatives of u improves itself with theorder of the derivatives. We will show that V = (π, u)T behaves similarly. Weshow that (ii) is true for ` ∈ N, ` ≥ 2, then by interpolation we obtain the resultfor all ` ∈ R, ` ≥ 2. Note that since m−1 > d/2, D2u0 and D2u are in L∞∩C0.

Proof of the proposition.

(i) Let V (x, t) = Du(x, t) and V0(x0) = Du0(x0). We have

V (X(x0, t), t) = (I + tV0(x0))−1V0(x0).

We write

V (X(x0, t), t) =1

(1 + t)I +

1

(1 + t)2K(x0, t),

where K(x0, t) = (1 + t)2(I + tV0)−1V0 − (1 + t)I. Then K is bounded on everycompact subset of Rd × R+. We now show that K stays bounded for t and x0

large.

Remark 1. Thanks to (H3), we show that:

• there exists a constant K = K(δ, |V0|∞) such that |V −10 |∞ ≤ K,

• there exists a constant L = L(δ, |V0|∞) such that |(I+tV0)−1|∞ ≤ L/(1+t).

1404 M. Grassin

To prove that, consider

V −10 =

1

det(V0)(adjV0)T ,

where adjV0 stands for the matrix of the cofactors of V0. With this formula, itis easy to see that |V −1

0 |∞ ≤ Cδ−d|V0|d−1

∞ .For the second point, we notice first that if µ is an eigenvalue of (I + tV0),

then λ = (µ − 1)/t is an eigenvalue of V0, and that |1 + t|/|1 + λt| ≤ C forλ ∈ SpV0. We write the same formula as in the previous case for (I + tV0)−1 toconclude.

Then for t large enough, one has ‖t−1V −10 (x0)‖ < 1 for all x0, and

K(x0, t) =(1 + t)2

t(I + t−1V −1

0 )−1 − (1 + t)I

=(1 + t)2

t

(I−

V −10

t+O

(1

t2

))− (1 + t)I

=(1 + t)

tI−

(1 + t)2

t2V −1

0 +O

(1

t

).

This is bounded independently of (x0, t). This proves part (i) of Proposition 2.

(ii) We know u and its derivatives with respect to x0 on the curves X(x0, t).We just have to deduce from that the expression of the derivatives with respectto x. Let W (x0, t) = V (X(x0, t), t). By induction, we show that, for k ≥ 1:

Dkx0W = (I + tV0(x0))−1Λk(I + tV0(x0))−1,

where Λk is a sum of products of t(I + tV0(x0))−1 and (DjV0), j ∈ {1, . . ., k},appearing βj times with

∑j jβj = k.

Then we use:

Dkx0W (x0, t) =

k∑j=1

DjxV (X(x0, t), t)

( ∑1≤ki≤k

Dk1x0X ⊗ . . .⊗Dkj

x0X)

with∑ji=1 ki = k, and

Dx0X = I + tV0(x0),

D`x0X = tD`−1V0(x0), for ` ≥ 2.

By induction, we show that for all j ≥ 1:IH(j): Dj

xV is a sum of terms which are products in a certain orderof: (I + tV0(x0))−1, tI, or (I + tV0(x0)) and D`V0 appearing β` times, with


∑` `β` = j. Moreover, the L∞-norm of the terms with t is bounded by a con-

stant times (1 + t)−(j+2), and we have ‖DjxV (X(·, t), t)‖0 ≤ C(1 + t)−(j+2), with

C = C(δ, ‖u0‖X).Suppose IH(k-1), then:

Dkx0V (X(x0, t), t) =

(Dkx0W (x0, t)−

k−1∑j=1

DjxV( ∑

1≤ki≤k−1

Dk1x0X ⊗ . . .⊗Dkj

x0X))

◦(

(I + tV0(x0))−1)⊗k

.

In the right-hand side term, one has the norm of the following terms to estimate:

a) (I + tV0(x0))−1Λk(I + tV0(x0))−(k+1).

b) DjxV (I + tV0(x0))j−s

∏ki 6=1

tDki−1V0(I + tV0(x0))−k,

with∑ki 6=1

(ki − 1) = k − j.

This last term correspond to the term where s of the kj are distinct from 1.Thus, the factor (I + tV0), which corresponds to Dx0

X, appears j − s times.For each of these terms, we apply first the induction hypothesis and we

consider the L∞-norm in space for the terms with t and we use the remark tobound it. We use IH(j) for j ≤ k−1 to show that b) is a product of (DjV0(·, t))βj

with∑j jβj = k and of tI, (I+ tV0)−1, such that the the L∞-norm of the terms

with t is bounded by a constant times (1 + t)−(k+2). Then we find an upperbound in L2−norm for ∏

1≤j≤k

(DjV0(·, t))βj

with∑

j jβj = k, by using Gagliardo-Nirenberg inequality. To conclude, we

obtain the upper bound in IH(k) which depends on δ, |V0|∞, and ‖DkV0‖0 for1 ≤ k ≤ m, that is to say on δ and ‖u0‖X .

Finally, we have to make a change of variables to obtain:

‖DjxV (·, t)‖0 ≤ (1 + t)d/2‖Dj

xV (X(·, t), t)‖0.

This gives (ii) since DjxV = D2u.

(iii) Since m − 1 > d/2, we know that D2u0 ∈ L∞. Thus D2u ∈ L∞, andwe have:

DxV (X(x0, t), t) = −(I + tV0(x0))−1DV0(x0)(I + tV0(x0))−2.

Using the remark again, we obtain: |DxV (X(x0, t), t)|∞ = O((1 + t)−3).

1406 M. Grassin

1.1.3. Energy estimates. We have now a guess of the behaviour of thevelocity. Therefore, we introduce u in the equations in order to estimate (π,w =u− u). Moreover, we will use the dispersive properties of u described in Propo-sition 2 and also the fact that (π, u− u)(·, t) ∈ Hm(Rd). The system for (π,w)is: {

(∂t + w · ∇)π + C1πdiv(w) = −u · ∇π − C1πdiv(u),

(∂t + (w · ∇))w + C1π∇π = −(u · ∇)w − (w · ∇)u,(7)

We note U = (0, u)T , U = (π,w)T , and U is solution of the following system:

∂tU +

d∑α=1

Aα(U)∂αU = −B(DU, U)−d∑

α=1

uα∂αU,(8)

where

B(DU, U) =

(C1πdiv(u)

(w · ∇)u

).

The right-hand side term in (8) provide more precise estimates thanks to Propo-sition 2. Before we perform the calculus, we have to choose the spatial norm weestimate. We consider the semi-norm which appears naturally in the calculus:

Yk(t) =

(∫RdDkU(x, t) ·DkU(x, t) dx

)1/2

.(9)

We expect Yk for k = 0, . . ., m to behave more or less like ‖Dku‖0(t), that is tosay to decay in time with a rate depending on k. Therefore, instead of using theclassical norm in Hm(Rd), we introduce

Z(t) =

m∑k=0

(1 + t)γkYk(t),(10)

where γk is chosen such that each term of the sum has the same decay in t. Thuswe will obtain an efficient estimate on Z. We take γk = k + r − a, and

r =

1−

d

2if γ ≥ γc = 1 +

2

d,

γ − 1

2d−

d

2if 1 < γ < γc.

We will choose a later. It will ensure that our calculus gives us a good estimate.We emphasize that Z depends on a. Note also that the coefficients γk can benegative.

We apply Dk on (8) and we take the inner product with DkU . Then weintegrate on Rd to obtain:

1

2

d

dt

∫RdDkU(x, t) ·DkU(x, t) dx =

∫RdRk(U)(x, t) dx(11)

+

∫RdSk(U, U)(x, t) dx,


with (dropping x and t):

Rk(U) = −DkU ·(Dk(∑

α

Aα(U)∂αU)−∑α

Aα(U)∂αDkU)

(12)

+1

2

∑α

DkU · ∂αAα(U)DkU,

Sk(U, U) = −DkU ·Dk(B(DU, U)) +1

2

(∑α

∂αuα

)DkU ·DkU(13)

− DkU ·(Dk(∑

α

uα∂αU)−∑α

uα∂αDkU).

To obtain this form, we used that, at each time, U has a compact support. Weused also the symmetric form of the system. We now find an upper bound forthe terms in the right-hand side of the inequality. But we first isolate some termsand estimate them more precisely: we will compute exactly the terms where aderivative of order one of U appears. The main reason is that Du has not a sogood decay in t, so that we will use the sign of the terms which contain Du tocontrol them in the estimate.We show:

Proposition 3. There exist C ∈ R+ depending only on m, d, and C ′ ∈ R+

depending on m, d, δ, ‖u0‖X , such that:∣∣∣∣∫RdRk(U)(x, t) dx

∣∣∣∣ ≤ C|DU |∞Y 2k ,(14)

k + r

(1 + t)Y 2k +

∫RdSk(U, U)(x, t) dx ≤ C ′YkZ(1 + t)−γk−2.(15)

Before we prove this proposition, we recall some inequalities that we will usein the calculus.

Lemma 1.[Gagliardo-Nirenberg inequality] Let r > 0, 0 ≤ i ≤ r, and z ∈L∞ ∩Hr. Then ∂iz ∈ L2r/i and∣∣∂iz∣∣

2r/i≤ Ci,r

∣∣z∣∣1−i/r∞

∥∥Drz∥∥i/r

0.

For a proof of this lemma, see [7]. We apply this and use Sobolev inequalitiesto obtain:

Lemma 2. Let 0 < p < d/2 and 1/q = 12 − p/d. There exists C such that

for all z ∈ Hp, we have |z|q ≤ C‖Dpz‖0, with C depending on p, q, d.

We have also thanks to Sobolev inequalities:

1408 M. Grassin

Lemma 3. Let p > d/2 and z ∈ Hp(Rd). Then:∣∣z∣∣∞≤ C

∥∥z∥∥1−θ

0

∥∥Dpz∥∥θ

0, with θ = d/2p.

We deduce from this:

Lemma 4. Let β = −γ1 − d/2. Since U ∈ Hm(Rd) and m > 1 + d/2, onehas:

(i) |U |∞(t) ≤ C(1 + t)β+1Z(t).

(ii) |DU |∞(t) ≤ C(1 + t)βZ(t).

(iii) If m > 2 + d/2, then |D2U |∞(t) ≤ C(1 + t)β−1Z(t).

Proof of Proposition 3.

(i) Rk is a polynomial function in DU , . . ., DkU homogeneous in weight anddegree. Its weight is 2k+1. Remark thatRk is sum of terms like ∂kU∂Ù∂k+1−Ùfor 1 ≤ ` ≤ k. Here, ∂kU stands for one particular derivative of order k of onecomponent of U , for example for (∂kπ)/(∂xk−1

1 ∂x2). If k 6= 0, 1, we applyLemma 1 to ∂U , and we obtain:

|∂jU |pj ≤ Cj−1,k−1|DU |1−2/pj∞ ‖DkU‖

2/pj0 , for pj = 2

k − 1

j − 1.(16)

If ` 6= k, ` 6= 1, since 1/p` + 1/pk−`+1 = 12 , we have by Holder’s inequality:∫

|∂kU∂Ù∂k+1−Ù | ≤ ‖DkU‖0‖∂Ù∂k+1−Ù‖0

≤ ‖DkU‖0|∂Ù |p` |∂

k+1−Ù |pk−`+1

≤ C|DU |∞‖DkU‖20.

The constant C depends only on m. We show the same estimate in the othercases and we obtain (14).

(ii) We note first that Sk is a sum of terms which are product of two deriva-tives in U and one in U . We split Sk in two terms: S1

k where we put all theterms with a derivative of order one of U , and S2

k where we put the terms witha derivative of order at least two of U . Then we will study precisely S1

k and findits sign, and we will find an upper bound for S2

k.

1. A precise analysis of S1k gives this expression:∫

S1k = I1 + I2 + I3,


with

I1 = −

∫RdDkU ·B(DU,DkU),

I2 =1

2

∫Rd

∑α

∂αuαDkU ·DkU,

I3 = −

∫Rd

∑1≤β1≤...≤βk≤d

(∂β1...βkU ·

k∑i=1

∑α

∂βi uα∂α∂β1...βi−1βi+1...βkU).

Recall that

B(DU,DkU) =

(C1D

kπdiv(u)

(Dkw · ∇)u

).

Using the first part of Proposition 2,

Du(x, t) =1

(1 + t)I +

1

(1 + t)2K(x, t),

we may write:

I1 = −C1d

1 + t

∫Dkπ ·Dkπ −

1

1 + t

∫Dkw ·Dkw +R1,

I2 =d

2

1

1 + tYk

2 +R2,

I3 = −k

1 + tYk

2 +R3.

And the error terms verify:

|Rj | ≤K

(1 + t)2Yk

2 j = 1, 2, 3,

where K is a constant depending on m, δ, and ‖u0‖X .Now, we have:∫

S1k ≤

K

(1 + t)2Yk

2 −Ak

(1 + t)

∫Dkw ·Dkw −

Bk(1 + t)

∫Dkπ ·Dkπ,

where Ak = 1−

d

2+ k,

Bk =γ − 1

2d−

d

2+ k.

Since Yk ≤ Z(1 + t)−γk , we have

K

(1 + t)2Yk

2 ≤ KYkZ(1 + t)−γk−2.

1410 M. Grassin

We use that

Y 2k =

∫(Dkπ ·Dkπ +Dkw ·Dkw)

and we chooser = min(Ak, Bk)− k.

Note that we loose some accuracy in the calculus at this point. We obtain:

k + r

(1 + t)Y 2k +

∫RdS1k(U, U)(x, t) dx ≤ C ′YkZ(1 + t)−γk−2.

2. In S2k, we have terms of the form: Q(U) = ∂kU∂Ù∂k+1−Ù for 1 ≤ k ≤ m

and 2 ≤ ` ≤ k + 1. To estimate these terms, we use the fact that, here, thederivatives of U are at least of order two. Hence we know that these terms havea good decay in (1 + t) thanks to Proposition 2. When m > 2 + d/2, we justapply Lemma 1 to obtain an upper bound where the worst terms are |D2U |∞and |D2U |∞. Thanks to part (iii) of Proposition 2, we have a good enoughestimate for the first one. When 1+d/2 < m ≤ 2+d/2, we can not do this sinceD2U /∈ L∞. But we manage to make the worst terms in the upper bound to be‖DnU‖0 and ‖DnU‖0 with some n ≥ 2, n ∈ R. Thus we can obtain the sameresult.

Case m > 2 + d/2We state first a result we need in the following. We apply Lemma 1 to obtain:

Lemma 5. Let z ∈ Hm such that D2z ∈ L∞. Then we have for allk ∈ [4,m], for all i ∈ [2, k], ∂iz ∈ Lp with p = 2(k − 3)/(i− 2) and

|∂iz|p ≤ Ci,k|D2z|1−2/p∞ ‖Dk−1z‖2/p0 .

Now we must study different case according to the values of k and `.

(a) k > 3 and 2 ≤ ` ≤ k − 1.Applying the previous lemma to z = U with i = `, then to z = U with

i = k + 1− `, we find that:

|∂Ù |p ≤ C|D2U |1−2/p∞ ‖Dk−1U‖2/p0 with p = 2

k − 3

`− 2

|∂k+1−Ù |q ≤ C|D2U |1−2/q∞ ‖Dk−1U‖2/q0 with q = 2

k − 3

k − `− 1.

Note that 1/q + 1/p = 12 . Thus, we obtain:∫ ∣∣Q(U)

∣∣ ≤ ∥∥DkU∥∥

0

∣∣∂Ù ∣∣p

∣∣∂k+1−Ù∣∣q

≤ CYk∣∣D2U

∣∣1−2/p

∞

∥∥Dk−1U∥∥2/p

0

∣∣D2U∣∣1−2/q

∞

∥∥Dk−1U∥∥2/q

0.


Using that:

|D2U |∞ ≤ C(1 + t)−3,

‖Dk−1U‖0 ≤ C(1 + t)d/2−k,

|D2U |∞ ≤ C ′Z(1 + t)β−1,

‖Dk−1U‖0 ≤ C ′Z(1 + t)−γk−1 ,

where C depends on m, d, ‖u0‖X , δ, and C ′ depends on m, d, we deduce that:∫|Q(U)| ≤ CYkZ(1 + t)dk ,

with

dk = −3

(1−

2

p

)+

2

p

(d

2− k

)+

(1−

2

q

)(β − 1)− γk−1

2

q

= −32

q+

(1−

2

q

)(d

2− k

)+

(1−

2

q

)(−2− (r − a)−

d

2

)− (γk − 1)

2

q

= −32

q+

(1−

2

q

)(−k − (r − a))− 2

(1−

2

q

)− γk

2

q+

2

q

= −γk − 2.

(b) If ` = k and k ≥ 2, ` = k + 1 and k ≥ 1, or k = 3 and ` = 2, an easycomputation gives the same estimate.

Case 1 + d/2 < m ≤ 2 + d/2Here again, we have to consider several cases.

(a) 2 ≤ ` ≤ k − 1.Note that 2 ≤ ` ≤ k − 1 ≤ m− 1 occurs only if d ≥ 2. Hence, we suppose in

this case that d ≥ 2.Let

n =1

2

(d

2+ k + 1

).

Remark that n > 2 and n < m. Thus

1

2≤ n− ` <

d

2and

1

2≤ n− (k + 1− `) <

d

2.

We apply Lemma 2 to z = ∂Ù with p = n − `, and to z = ∂k+1−Ù withs = n− (k + 1− `). We get:

|∂Ù |q ≤ C‖DnU‖0, with 1/q = 1/2− (n− `)/d,

|∂k+1−Ù |s ≤ C‖DnU‖0, with 1/s = 1/2− (n− (k + 1− `))/d .

1412 M. Grassin

We notice that:1

q+

1

s= 1−

(2n− k − 1)

d=

1

2.

Finally, we get: ‖∂Ù∂k+1−Ù‖0 ≤ C|∂Ù |q|∂k+1−Ù |s, and∫Q(U)| ≤

∥∥DkU∥∥

0

∣∣∂Ù ∣∣q

∣∣∂k+1−Ù∣∣s≤ CYk

∥∥DnU∥∥

0

∥∥DnU∥∥

0

≤ CYkZ(1 + t)dk ,

where

dk =

(d

2− (n+ 1)

)− γn

= −2− γk.

We used Proposition 2 with n ≥ 2.

(b) If k ≥ 2 and ` = k, or k ≥ 1 and ` = k + 1, we have easily the sameestimate.

This finishes the proof of the part (ii) of the estimate since one has in everycase: ∣∣∣∣∫ S2

k

∣∣∣∣ ≤ C ′ZYk(1 + t)−γk−2.

with C depending only on m, d, ‖u0‖X , and δ.

1.1.4. Conclusion. Now we have with (11) and Proposition 3:

1

2

d

dtYk

2 +k + r

(1 + t)Yk

2 ≤ C|DU |∞Yk2 + C ′YkZ(1 + t)−γk−2.(17)

We simplify by Yk, multiply by (1 + t)γk , and sum over k to obtain:

dZ

dt(t) +

a

(1 + t)Z(t) ≤ C|DU |∞Z(t) +

C ′

(1 + t)2Z(t).(18)

The constants C, C ′ are positive, C depends only on γ, m, d, and C ′ dependson γ, m, d, δ, ‖u0‖X . Since |DU |∞ ≤ C(1 + t)βZ, we choose β = 0 to obtain agood estimate, and this leads to

a = 1 + r +d

2> 1.

Thus:dZ

dt(t) +

a

(1 + t)Z(t) ≤ C(Z(t))2 +

C ′

(1 + t)2Z(t).(19)

To conclude, we use the following simple result:


Proposition 4. Since a > 1, there exists Λ = Λ(a,m, d, δ, ‖u0‖X) > 0 suchthat the Cauchy problem:

dZ

dt+

a

(1 + t)Z = CZ2 +

C ′

(1 + t)2Z,

Z(0) = Z0 < Λ,

(20)

has a global solution for t ≥ 0.

Proof. We claim that the solution of this differential equation is:

Z(t) =

(1 + t)−a exp

(C ′(

1−1

1 + t

))(

1

Z0

−

∫ t

0

C(1 + τ )−aeC′(1−1/(1+τ))dτ

) .Thus, Z is defined for t ≥ 0 if and only if:

0 < Z0 < Λ =1∫ ∞

0

C(1 + τ )−aeC′

e−C′1/(1+τ)dτ

.

This condition can be filled only if the integral converges, that is to say only ifa > 1. Note that if a ≤ 1, there is no global solution to the differential equationwe consider. In our case, it suffices to choose Z(0) small enough to satisfiesthe condition. Then we have a global solution Z which moreover satisfies theestimate:

Z(t) ≤ K(1 + t)−a for all t ≥ 0,

where K depends only on m, d, δ, ‖u0‖X , Z0.

Finally, if Z(0) is small enough, Z(t) is less than the global solution Z givenby Proposition 4. The condition on Z(0) is: Z(0) < Λ. This corresponds tohypothesis (H1) since we have:

Z(0) =

m∑k=0

∥∥Dk(ρ(γ−1)/20 )

∥∥0

=∥∥ρ(γ−1)/2

0

∥∥m.

We have obtained the estimate:

Z(t) ≤ K(1 + t)−a for all t ≥ 0,

where K depends only on m, d, ‖u0‖X , δ, and ‖ρ(γ−1)/20 ‖m. Then, we deduce

thatYk(t) ≤ K(1 + t)−(k+r) for all t ≥ 0.

These estimates lead to Theorem 2.We can conclude that the solution is global since the L2-norm of its derivatives

are bounded by functions of t which never blow up.

1414 M. Grassin

1.2. Global in time uniqueness. In this part, we state a result of localin space and global in time uniqueness. Then, we prove a corollary of Theorem1 in which we get rid of hypothesis (H4).

We now show a result of global uniqueness in time. We compare the globalsolution of Theorem 1 and another solution with enough regularity.

We note π0 = ρ(γ−1)/20 .

Proposition 5. Let (ρ0, u0) satisfy (H1)-(H4). Let U = (π, u)T be theglobal solution of (5) given by Theorem 1 and u be the solution of (4). Consider Va global solution of (5) such that DV ∈ L∞(Rd×R+). Then, for all ν ∈ ]2−a, 1[,for all R0 > 0, there exists T0 > 0 such that, if U(·, T0) = V (·, T0) on B(0, R0),then U and V are equal on the domain {(x, t) : |x−x(t)| ≤ R(t), for all t ≥ T0},where x(t) is the solution of x′(t) = u(x(t), t), x(T0) = 0, and R(t) = R0(1+ t)ν .

x(t)

BT

D

BT0

Figure 2: Domain D

We note Bt = B(x(t), R(t)). For a T0 given big enough, we show now thatif U(·, T0) = V (·, T0) on B(0, R0), then U(·, t) = V (·, t) on Bt with R(t) =R0(1 + t)ν , ν < 1, for all t ≥ T0. The choice of T0 depends on ν, R0, a, ‖u0‖X ,

δ and ‖ρ(γ−1)/20 ‖m.


We prove now that there exists T0, x(t) and R(t) such that U = (π, u)T andV = (π, v)T solutions of (5) satisfy U = V on D = {(x, t)/|x− x(t)| ≤ R(t), forall t ≥ T0}. For that, we consider T > T0 and we evaluate the norm:

|U − V |2BT =

∫BT

(U − V ) · (U − V )(x, T ) dx.

We write the equation satisfied by U − V on D:

∂t(U − V ) +

d∑α=1

(Aα(U)∂αU −Aα(V )∂αV ) = 0.

It is also:

∂t(U − V ) +

d∑α=1

Aα(U)(∂αU − ∂αV ) = −d∑

α=1

(Aα(U)−Aα(V ))∂αV.

Then, we take the inner product with U − V and integrate over

DT = {(x, t) : |x− x(t)| ≤ R(t), T0 ≤ t ≤ T} :

∫DT

1

2∂t(|U − V |

2) +1

2

d∑α=1

∂α((U − V ) ·Aα(U)(U − V ))

=

∫DT

1

2

d∑α=1

(U − V ) · ∂α(Aα(U))(U − V )

−

∫DT

d∑α=1

(U − V ) · (Aα(U)−Aα(V ))∂αV .

By Stokes’ formula, we get:

1

2

∫∂DT

|U − V |2nt +

d∑α=1

(U − V ) ·Aα(U)(U − V )nα

=1

2

∫DT

d∑α=1

(U − V ) · ∂α(Aα(U))(U − V )

−

∫DT

d∑α=1

(U − V ) · (Aα(U)−Aα(V ))∂αV ,

where n = (nt, n1, . . . , nd) is the normal vector to DT .

1416 M. Grassin

This gives:

1

2

∫∂DT

|U − V |2nt +d∑

α=1

(U − V ) ·Aα(U)(U − V )nα

≤ C

∫ T

T0

(|DU |L∞(D) + |DV |L∞(D)

)∣∣U − V ∣∣2Bt.

To express the first term, we need to precise n and ∂DT . We have:

∂DT = ({T0} ×BT0) ∪ ({T} ×BT ) ∪M

where

M = {(t, x) | x = γ(t) = x(t) +R(t)y, for y ∈ S(0, 1) and T0 ≤ t ≤ T}.

On {T0}×BT0we have n = (−1, 0, . . . , 0), on {T}×BT we have n = (1, 0, . . . , 0),

and on M,

n =1√

1 +∣∣∣y · ∂γ∂t ∣∣∣2

(−y ·

∂γ

∂t, y1, . . . , yd

).

We deduce from that the expression of the left-hand side term:

1

2

∫∂DT

|U − V |2nt +d∑

α=1

(U − V ) ·Aα(U)(U − V )nα

=1

2(|U − V |2BT − |U − V |

2BT0

) +1

2

∫M

1√1 +

∣∣∣y · ∂γ∂t ∣∣∣2(−|U − V |2y ·

∂γ

∂t

+d∑

α=1

(U − V ) · Aα(U)(U − V )yα

)dσ.

We note

Ψ = −|U − V |2y ·∂γ

∂t+

d∑α=1

(U − V ) ·Aα(U)(U − V )yα,

and using the expression of Aα(U), we get:

Ψ = |U − V |2[(u(x(t) +R(t)y, t)− x′(t)) · y −R′(t)

]+ 2C1π(π − π)y · (u− v).

We claim that our choice of x and R implies that there exists T0 such that:

Ψ > 0, for all t ≥ T0, for all y ∈ S.


Indeed, we know that Du(x, t) = Du(x, t) +Dw(x, t), where

|Dw(·, t)|∞ ≤ K(1 + t)−a,

with K a constant depending on ‖u0‖X , δ, and ‖ρ(γ−1)/20 ‖m. Therefore

Du(x, t) =1

1 + tI +O((1 + t)−a) .

Using that x′(t) = u(x(t), t), we integrate to obtain:

(u(x(t) +R(t)y, t)− x′(t)) · y =R(t)

1 + t+O

(R(t)

(1 + t)a

),

where the notation

O

(R(t)

(1 + t)a

)means that this function is bounded for all x by K

(R(t)/(1 + t)a

), where K is

independent of t, x.We estimate the last term by:

|2C1π(π − π)y · (u− v)| ≤ C1|π|∞(|π − π|2 + |u− v|2)

≤ KC1(1 + t)1−a|U − V |2.

Thus we obtain that:

Ψ = |U − V |2(R(t)

1 + t−R′(t) +O

(R(t)

(1 + t)a

)+O

((1 + t)1−a

)).

Thus our claim is true if R(t) = R0(1 + t)ν , for ν < 1, and t big enough. Indeed,we have then

R(t)

1 + t−R′(t) = R0(1− ν)(1 + t)ν−1,

KR(t)

(1 + t)a= KR0(1 + t)ν−a.

Hence, we choose T0 such that for t ≥ T0, we have:

KR(t)(1 + t)−a + KC1(1 + t)1−a <R(t)

1 + t−R′(t),

orK

1− ν(1 + t)1−a + C1

K

R0(1− ν)(1 + t)2−ν−a < 1.

There exists T0 such that this is true for t ≥ T0 if ν ∈ ]2−a, 1[.

1418 M. Grassin

The size of T0 depends of K(‖u0‖X , δ, ‖ρ(γ−1)/20 ‖m), R0, ν, and a: the smaller

ν is and the bigger a is, the smaller T0 is.Now we have for this choice of T0, x, and R:

1

2(|U − V |2BT − |U − V |

2BT0

) ≤ C

∫ T

T0

(|DU |∞(t) + |DV |∞(t))|U − V |2Btdt.

And we deduce that for all T > T0:

1

2(|U − V |2BT − |U − V |

2B0

) ≤ C

∫ T

T0

(|DU |L∞(D) + |DV |L∞(D))|U − V |2Btdt.

Now we use the fact that |DU |∞(t) ≤ K(1 + t)−1 as we have shown in the firstsection. And we use also the hypothesis on DV to conclude that:

1

2(|U − V |2BT − |U − V |

2BT0

) ≤ C

∫ T

T0

|U − V |2Btdt,

where C depends on ‖u0‖X , δ, ‖ρ(γ−1)/20 ‖m, |DV |∞. Then we conclude easily by

a Gronwall’s inequality, since U(·, T0) = V (·, T0) on BT0, that U(·, T ) = V (·, T )

in BT for all T > T0.

1.3. A corollary. Now we prove a corollary of Theorem 1 which improvesslightly the hypotheses.

Corollary 1. Let (ρ0, u0) be the initial data for Problem (2). Suppose(H1)-(H3). Then the result of Theorem 1 is still true.

Proof. We aim at defining a solution for ρ0 satisfying only (H1). For thatwe will use the result in the compact support case. We start by introducing afunction ψ ∈ C∞c (Rd,R) such that ψ ≡ 1 on B(0, 1). Then we take: ρk0(x) =ρ0(x)ψ(x/k) for k ∈ N. We claim that:

• ρk0 ≡ ρ0 on B(0, k) and ρk0 has a compact support.

• πk0 = (ρk0)(γ−1)/2 ∈ Hm(Rd) and ‖πk0‖m ≤ C‖ψ(γ−1)/2‖m‖ρ(γ−1)/20 ‖m,

where C is independent of k.

Thus we can apply Theorem 1 to ρk0 noticing that the smallness of (ρk0)(γ−1)/2

does not depend on k. We obtain Uk = (πk, uk)T , a solution associated toUk0 = (πk0 , u0)T . We have πk0 ≡ π

k+10 on B(0, k).

We apply the result of local in space and time uniqueness. The solutions arethe same on {(x, t) | 0 ≤ t ≤ T0, |x| ≤ k −Mkt}, provided that the initial dataare the same on B(0, k), where

Mk ≥ sup0≤t≤T0

(C1|π

k|L∞(B(0,k))(t) + |uk|L∞(B(0,k))(t)).


Now we have, thanks to our estimates:

|πk|∞(t) ≤ C,

|uk|L∞(B(0,k))(t) ≤ C + |u|L∞(B(0,k))(t),

where C depends on δ, ‖u0‖X , ‖π0‖m.Then we estimate |u|L∞(B(0,k)). We choose T0 = 1/(2|Du0|∞) to obtain the

following lemma:

Lemma 6. If (x, t) ∈ [0, T0] × B(0, k) and x = x0 + tu0(x0), then x0 ∈B(0, Rk) with Rk = 2(k + T0|u0(0)|).

We deduce from this lemma the estimate:

|u|L∞(B(0,k)) ≤ |Du0|∞Rk + |u0(0)|.

Then the condition on Mk is:

Mk ≥ 2C + 2|u0(0)|+ 2|Du0|∞k.

Thus, for k big enough, we can take Mk ≤ ck, where

c = c(∣∣u0(0)

∣∣, ∣∣Du0

∣∣∞, δ,∥∥u0

∥∥X,∥∥π0

∥∥m

).

Now, we consider T < min(T0, 1/(2c)). Then, we know that Uk = Uk+1 on{(x, t) | 0 ≤ t ≤ T0, |x| ≤ k −Mkt}. Note that k −MkT ≥ k − ckT ≥ k/2.Therefore, we have shown that Uk = Uk+1 on B(0, k/2)× [0, T ].

We can now define

U(x, t) = (π(x, t), u(x, t))T

by

Uk = (πk, uk)T on B

(0,k

2

)× [0, T ]

for k big enough. We know that U − (0, u)T is in C0([0, T ];Hm(B(0, R))) for allR ∈ R+, by using the property of each Uk. We have to show that U − (0, u)T ∈C0(0, T ;Hm(Rd)), and thus U will be a local solution of our problem. Moreover,we will show that U satisfies the same energy estimates as Uk, so that U is aglobal solution. We set V = U − (0, u)T . We have:∥∥DjV

∥∥2

0(t) = supR>0

∥∥DjV∥∥2

L2(B(0,R))(t).

Hence for k big enough:∥∥DjV∥∥2

L2(B(0,R))≤

∥∥Dj(Uk − (0, u)T )∥∥2

0(t)

≤ C(1 + t)−2(a+γj),

1420 M. Grassin

since ∥∥Dj(Uk − (0, u)T )∥∥2

0(t) ≤ C(1 + t)−2(a+γj),

and C only depends on m, d, δ, ‖u0‖X , ‖ρ(γ−1)/20 ‖m.

We conclude that (U − u)(t) ∈ Hm(Rd) and that

Z(t) =

m∑j=1

(1 + t)γj∥∥DjV

∥∥0

satisfies:Z(t) ≤ C(1 + t)−a.

In the same way, we verify that

U − (0, u)T ∈ Cj([0,∞[;Hm−j(Rd)) for j = 0, 1.

2. The non-isentropic case. The results in the general case are mainlythe same as in the isentropic one. Nevertheless, some difficulties, especially inthe estimates, appear due to the presence of the entropy in the equations.

We express a result of global existence of a smooth solution for (1), and westate some properties of uniqueness which can be proved as in the isentropiccase. Recall the system that we consider with (ρ0, u0, S0) as initial data:

∂tρ+ div(ρu) = 0,

ρ(∂tu+ (u · ∇)u) +∇p = 0,

∂tS + u · ∇S = 0,

and the pressure law is the following: p = (γ − 1)ρe.

2.1. Global existence. We prove the following result:

Theorem 3. Let m > 1 + d/2. Suppose that

(H1) (ρ(γ−1)/20 , S0) is small enough in Hm(Rd),

(H2) u0 ∈ X,(H3) there exists δ > 0 such that for all x ∈ Rd, dist(Sp (Du0(x)),R−) ≥ δ,(H4) ρ0 and S0 have a compact support.

Let u be the global solution to (4). Then there exists a global smooth solution tothe Cauchy problem for (1), i.e., (ρ, u, S) such that

(ρ(γ−1)/2, u− u, S) ∈ Cj([0,∞[;Hm−j(Rd))for j ∈ {0, 1}.

In (H3), dist stands for the distance and Sp for the spectrum. Note thatwe can suppose that S0 − S0 is small in Hm(Rd) where S0 is a constant. Thescheme of the proof is the same as in the isentropic case. But we have to studymore precisely the terms appearing in Rk and Sk, and to adapt our norm sincewe have a new term with S.

We obtain the following estimates:


Theorem 4. Under the hypotheses (H1)-(H4), the solution of Theorem 3satisfies:

• forall 1 ≤ k ≤ m, all t ≥ 0, ‖DkT‖0(t) ≤ K(1 + t)−k−b/2+d/2,

forall 1 ≤ k ≤ m, all t ≥ 0, ‖DkS‖0(t) ≤ K(1 + t)−k+d/2,

• forall t ≥ 0, |S|∞(t) ≤ K,forall t ≥ 0, |T |∞(t) ≤ K(1 + t)−b/2,

• forall t ≥ 0, |DT |∞(t) ≤ K(1 + t)−1−b/2,forall t ≥ 0, |DS|∞(t) ≤ K(1 + t)−1.

where T stands for π or w = u−u, K depends on δ, ‖u0‖X , ‖ρ(γ−1)/20 ‖m, ‖S0‖m,

and b is a real number such that

0 < b < min

(1,γ − 1

2d

).

2.1.1. Local existence. We have first to symmetrize the system in takinginto account S. As in [6], we take

π =

(γ − 1

2

p

γ − 1

)(γ−1)/2γ

,

and we obtain the following system from (1):eS/γ(∂t + u · ∇)π + C1e

S/γπdiv(u) = 0 ,

(∂t + (u · ∇))u+ C1eS/γπ∇π = 0 ,

∂tS + u · ∇S = 0 .

(21)

The construction of a local solution is the same as in the isentropic case. LetR > 0 be such that suppρ0 ⊂ B(0, R) and suppS0 ⊂ B(0, R). Let ϕ ∈ C∞c (Rd)be such that ϕ ≡ 1 on B(0, R + 2η) where η > 0. We consider the sameapproximate problem on the velocity (4) and we use the result of Propositions2. We use also the general theorem of local existence of smooth solution forsymmetric hyperbolic systems —cf [2]— to obtain a local solution (πϕ, uϕ, Sϕ)with (π0, u0ϕ, S0) ∈ Hm as initial data. A result of local uniqueness can beproved as in the isentropic case -see in the next section-. With this result, weobtain (π, u, S) a solution of (21) defined by:{

(πϕ, uϕ, Sϕ) on K,

(0, u, 0) outside K,

where

K = {(x, t) | 0 ≤ t ≤ T , x ∈ B(0, R+ η +Mt)} with:

1422 M. Grassin

M = sup0≤t≤Tex−ε

(C1e|S0|∞/2γ |πϕ|L∞ + |uϕ|L∞), and

T = min(Tex − ε,

η

2M− ε), for ε > 0 given.

And by construction (π, u− u, S) ∈ Cj([0,∞[;Hm−j(Rd)) for j ∈ {0, 1} and hasa compact support.

2.1.2. Estimates in the general case. Thus, we compare V and U =(0, u, 0)T and we note w = u − u , U = (π,w, S)T . Using the fact that theequation on S is a simple transport equation, which is not deeply linked to therest of the system, we put a weight in t on S to control its role in the estimates.We write the system in the following way:

eS/γ(∂t + w · ∇)π + C1eS/γπdiv(w) = −eS/γ(u · ∇π − C1πdiv(u)),

(∂t + (w · ∇))w + C1eS/γπ∇π = −(w · ∇)u− (u · ∇)w,

(1 + t)−b∂tS + w · ∇((1 + t)−bS) = −u · ∇((1 + t)−bS),

where b is a real number that we will choose later. We write this also:

A0(t, U)∂tU +

d∑α=1

Aα(U)∂αU = −B(DU, U)−d∑

α=1

Cα(U)∂αU,(22)

where A0(t, U) = diag(eS/γ , 1, . . . , 1, (1 + t)−b) ∈Md+2(R) is symmetric positivedefinite—we denote it by A0 in the following—. Each Aα(V ) ∈ Md+2(R) issymmetric:

Aα(V ) =

eS/γuα 0 . . . C1eS/γπ . . . 0 0

0 uα 0...

...... 0

. . .

C1eS/γπ

.... . .

......

. . . 0...

...... . . . 0 uα 0

0 0 . . . . . . 0 (1 + t)−buα

,

Cα(U) = uαA0 ∈Md+2(R) is symmetric, and

B(DU, U) =

C1e

S/γπdiv(u)

(w · ∇)u

0

∈Md+2(R).


Note that S is bounded in L∞-norm since it is the solution of a transportequation and S0 is in L∞(Rd). We choose the semi-norm:

Yk(t) =

(∫RdDkU(x, t) ·A0(t, U)DkU(x, t) dx

)1/2

.

Then, we introduce

Z(t) =

m∑k=0

(1 + t)γkYk(t),

where γk is chosen such that each term of the sum has the same decay in t. Infact, we choose:

γk = k + r − a, r =b

2−d

2, with 0 < b ≤ min

(1,γ − 1

2d

).

We determine a later on. As in the isentropic case, the expression k+ r appearsin the calculus when we study the role of DU in the estimates, i.e., when wecompute S1

k.We applyDk on (22), we take the inner product withDkU . Then we integrate

on Rd and we use Stokes’ formula. We obtain:

1

2

d

dt(Yk)2 =

∫RdRk(U)(x, t) dx+

∫RdSk(U, U)(x, t) dx,(23)

with -dropping x and t-:

Rk(U) = −∑α

DkU ·A0(Dk((A0)−1Aα(U)∂αU)− (A0)−1Aα(U)∂αDkU)

+1

2

∑α

DkU · ∂αAα(U)DkU −

1

2γeS/γDkπ · (w · ∇S)Dkπ,

Sk(U, U) = −DkU ·A0Dk((A0)−1B(DU, U)) +1

2

∑α

DkU · ∂αCα(U)DkU

−∑α

DkU ·A0(Dk((A0)−1Cα(U)∂αU)− (A0)−1Cα(U)∂αDkU)

−1

2γeS/γDkπ · (u · ∇S)Dkπ −

b

2(1 + t)−b−1DkS ·DkS.

We show the estimates:

Proposition 6. There exist C ∈ R+ depending only on m, d, and C ′ ∈ R+

depending on m, d, δ, ‖u0‖X , such that:∣∣∣∣ ∫RdRk(U)(x, t) dx

∣∣∣∣ ≤ C ∑x∈Ek

YkZ2+x(1 + t)e(x),(24)

1424 M. Grassin

where Ek = {0, 1, 2, . . . , k, p/(k − 1) for 1 ≤ p ≤ k − 1} and e(x) = −γk + β +(β + 1 + b/2)x = −γk + β + ax.

k + r

(1 + t)Y 2k +

∫RdSk(U, U)(x, t) dx ≤ C ′YkZ(1 + t)−γk−2.(25)

Recall that

β = −γ1 −d

2= −1−

b

2+ a.

The main difference between this proposition and the Proposition 3 standsin the point (i) and comes from the term eS/γ in our matrices. We first showthat Sk is not modified by this term and then we find an upper bound for thenew terms appearing in Rk because of this factor eS/γ .

Proof.(i) First we consider Sk and we make a cancellation in its terms. We have

indeed:

∂αCα(U) =

eS/γ

(1γ

(∂αS)uα + ∂αuα

)0 0

0(∂αuα I

)0

0 0 (1 + t)−b∂αuα

.

Thus:

1

2

d∑α=1

DkU · ∂αCα(U)DkU =

1

2γeS/γDkπ · (u · ∇S)Dkπ

+1

2

∑α

∂αuα(DkU ·A0DkU).

Therefore we have:

Sk = −DkU ·A0Dk((A0)−1B(DU, U)) +1

2divu(DkU ·A0DkU)

−∑α

DkU ·A0(Dk((A0)−1Cα(U)∂αU)− (A0)−1Cα(U)∂αDkU)

−b

2(1 + t)−b−1DkS ·DkS.

Now, we note that (A0)−1 and Cα(U) are two diagonal matrices and that theproduct (A0)−1Cα(U) equals to uαI. Remark also the simple expression of theproduct (A0)−1B(DU, U) = B(DU, U) with

B(DU, U) =

C1πdiv(u)

(w · ∇)u

0

∈Md+2(R).


As a consequence, we claim that the calculus for Sk are the same as in theisentropic case, except for the presence of a term DkS ·DkS. We use here thatS is bounded in L∞-norm by |S0|∞ to find an upper bound for eS/γ . As aconsequence, we have:

Sk = −DkU ·A0Dk(BU) +1

2divu(DkU ·A0DkU)

−∑α

DkU ·A0(Dk(uα∂αU)− uα∂αDkU)

−b

2(1 + t)−b−1DkS ·DkS.

We note Sk = S1k + S2

k, where S1k contains the terms with a derivative of

order one of U , and S2k contains the terms with a derivative of order at least two

of U . We have:∫S1k = −

∫RdDkU ·A0B(DU,DkU)−

∫Rd

b

2(1 + t)−b−1DkS ·DkS

−

∫Rd

∑1≤β1≤...≤βk≤d

(∂β1...βk

U ·A0k∑i=1

∑α

(∂βi uα)∂α∂β1...βi−1βi+1...βkU)

+1

2

∫Rd

divu(DkU ·A0DkU).

Recall Proposition 2 and replace Du in the previous expression:∫S1k =

∫Rd

d2 − k

1 + tDkU ·A0DkU −

∫Rd

C1d

1 + t(eS/γDkπ ·Dkπ)

−

∫Rd

1

1 + tDkw ·Dkw −

∫Rd

b

2(1 + t)(1 + t)−bDkS ·DkS

+

∫RdO

(1

(1 + t)2

)DkU ·A0DkU.

Now we use the definition of Yk to write this:∫S1k =

d2 − k

1 + tY 2k −

1

1 + t

(γ − 1

2d

∫RdeS/γDkπ ·Dkπ

+

∫RdDkw ·Dkw +

b

2

∫Rd

(1 + t)−bDkS ·DkS

)+O

(1

(1 + t)2Y 2k

).

We have

r =b

2−d

2= min

(γ − 1

2d, 1,

b

2

)−d

2,

1426 M. Grassin

since we have already chosen b such that

b

2≤ min

(1,γ − 1

2d

).

Then we obtain:∫S1k +

k + r

1 + tY 2k ≤

C

(1 + t)2Y 2k ≤

C

(1 + t)2YkZ(1 + t)−γk .

And for S2k there is no difference from the isentropic case. It is a sum of

terms like

Q(T ) = ∂kT∂`U∂k+1−`T

for 1 ≤ k ≤ m, 2 ≤ ` ≤ k + 1, and T = π or T = w. Thus we can show:∣∣∣∣∫ S2k

∣∣∣∣ ≤ C

(1 + t)2YkZ(1 + t)−γk .

Then we have proved the estimate (25).

(ii) Now we consider Rk. We note that the product (A0)−1Aα(U) containssome terms with eS/γ . As a consequence, when we compute

Dk((A0)−1Aα(U)∂αU)

some new terms arise in addition of those we had in the isentropic case. Theyare of the form:

Q(U) = ∂kT

( p∏j=1

(∂jS)βj)∂`−pT∂k+1−`T,

with 1 ≤ ` ≤ k, 1 ≤ p ≤ `, and∑j jβj = p, for T = π or T = w. We have also

the usual terms and the terms corresponding to S:{Q′(U) = ∂kT∂`T∂k+1−`T for 1 ≤ ` ≤ k and T = π or T = w,

Q′(U) = (1 + t)−b∂kS∂`T∂k−`+1S for 1 ≤ ` ≤ k.

For these last terms, we find an upper bound as in the Proposition (i) of 3, thatis to say: ∫

|Q′(U)| ≤ C|DU |∞Y2k ≤ CYkZ

2(1 + t)θk ,

with θk = −γk + β. This corresponds to the case x = 0 in the proposition. Nowwe have to find an upper bound for the terms like Q(U). We consider four cases.

(a) 1 ≤ p ≤ `− 1. Note that in this case we have necessarily k ≥ 2.


Let

pj = 2k − 1

j, q = 2

k − 1

`− p− 1, s = 2

k − 1

k − `.

We havep∑j=1

βjpj

+1

q+

1

s=

1

2.

Then: ∫ ∣∣Q(U)∣∣ ≤ ∥∥∂kT∥∥

0

∥∥∥∥ p∏j=1

(∂jS)βj∂`−pT∂k+1−`T

∥∥∥∥0

,

≤∥∥∂kT∥∥

0

p∏j=1

∣∣∂jS∣∣βjpj

∣∣∂`−pT ∣∣q

∣∣∂k+1−`T∣∣s.

We use Lemma 1 ∣∣∂jS∣∣pj≤ C

∣∣S∣∣1−2/pj

∞

∥∥Dk−1S∥∥2/pj

0,∣∣∂`−pT ∣∣

q≤ C

∣∣DT ∣∣1−2/q

∞

∥∥DkT∥∥2/q

0,∣∣∂k+1−`T

∣∣s≤ C

∣∣DT ∣∣1−2/s

∞

∥∥DkT∥∥2/s

0.

Then since T = π or T = w, ‖DkT‖0 ≤ CYk, and going back to the definition ofYk, we see that ‖Dk−1S‖0 ≤ C(1 + t)b/2Yk−1. The same remarks can be madefor the L∞-norm

|S|∞ ≤ C,

|DT |∞ ≤ CZ(1 + t)β, for T = π or T = w,

|DS|∞ ≤ CZ(1 + t)β+b/2.

Thus, we obtain:∫|Q(U)| ≤ CYk

∣∣S∣∣∑j(1−2/pj)βj

∞

∥∥Dk−1S∥∥2∑

jβj/pj

0

∣∣DT ∣∣2−2/q−2/s

∞

∥∥DkT∥∥2/q+2/s

0

≤ CYkZ2+p/(k−1)(1 + t)θk ,

with

θk =

(−γk−1 +

b

2

)(2∑j

βjpj

)+ β

(2−

2

q−

2

s

)− γk

(2

q+

2

s

)

=

(−γk + 1 +

b

2

)(2∑j

βjpj

)+ β

(1 + 2

∑j

βjpj

)− γk

(2

q+

2

s

)=

1428 M. Grassin

= −γk + β +

(1 +

b

2+ β

)(2∑j

βjpj

)

= −γk + β +

(1 +

b

2+ β

)p

k − 1.

This gives x = p/(k − 1) for 1 ≤ p ≤ `− 1 in the proposition.

(b) 1 ≤ p = ` and∑βj = 1. Note that 1 ≤ p =

∑j jβj . Thus we have nec-

essarily β` = 1 and βj = 0 for j 6= `. That is to say Q(U) = ∂kT∂`ST∂k+1−`T .Let

q = 2k − 1

`− 1, s = 2

k − 1

k − `.

We have ∫ ∣∣Q(U)∣∣ ≤ ∥∥∂kT∥∥

0

∣∣T ∣∣∞

∣∣∂`S∣∣q

∣∣∂k+1−`T∣∣s.

We use that ∣∣T ∣∣∞≤ Z(1 + t)β+1,∣∣∂lS∣∣

q≤ C

∣∣DS∣∣1−2/q

∞

∥∥DkS∥∥2/q

0,∣∣∂k+1−`T

∣∣s≤ C

∣∣DT ∣∣1−2/s

∞

∥∥DkT∥∥2/s

0.

Then we obtain easily ∫|Q(U)| ≤ CYkZ

3(1 + t)θk ,

with θk = −γk+β+(1 + b/2 + β). This gives the value x = 1 in the proposition.

(c) 1 ≤ p = ` and∑βj = k. This case occurs only when β1 = k and βj = 0

for j 6= 1. That is to say Q(U) = ∂kT (∂S)kT∂T . We have:∫|Q(U)| ≤ CYk|DS|

k∞ ‖T‖0 |DT |∞,

and this leads easily to ∫|Q(U)| ≤ CYkZ

2+k(1 + t)θk ,

with θk = −γk + β + (1 + b/2 + β) k. This gives x = k in the proposition.

(d) p = ` and 1 <∑βj < k. We set

pj = 2k − 1

j − 1, s = 2

k − 1

k − `,

1

q=

1

2−

(1

s+∑ βj

pj

).


Thus we have

2 < q =2(k − 1)∑βj − 1

and ∫Q(U) ≤

∥∥∂kT∥∥0

∥∥∥∥( ∏j=1

(∂jS)βj)T∂k+1−`T

∥∥∥∥0

≤∥∥∂kT∥∥

0

∏j=1

∣∣∂jS∣∣βjpj

∣∣T ∣∣q

∣∣∂k+1−`T∣∣s.

We use Lemma 1:

|∂jS|pj ≤ C|DS|1−2/pj∞ ‖DkS‖

2/pj0 ,

|∂k+1−`T |s ≤ C|DT |1−2/s∞ ‖DkT‖2/s0 .

And we use the Sobolev’s imbeddings theorem to find an upper bound to |T |q.Indeed, for n = d/2 and for all q ≥ 2, Hn(Rd) ⊂ Lq(Rd). We deduce from thatthe following estimate:

|T |q ≤ CZ(1 + t)−γ0−d(1/2−1/q).

Thus we obtain ∫|Q(U)| ≤ CYkZ

2+∑

βj (1 + t)θk ,

with θk = −γk + β + (1 + b/2 + β)∑βj . This gives x = 2, · · ·, k − 1 in the

proposition.

2.1.3. Conclusion. Now we conclude using (23) and Proposition 6. Wehave proved that:

1

2

d

dtYk

2 +k + r

(1 + t)Yk

2 ≤C

(1 + t)2YkZ(1 + t)−γk(26)

+ C∑x∈Ek

YkZ2+x(1 + t)e(x),

with Ek = {0, p/(k − 1) for 1 ≤ p ≤ k − 1, 1, 2, . . . , k} and e(x) = −γk + β +(β + 1 + b/2)x.

Recall that γk = k + r − a and r = b/2 − d/2. We simplify by Yk, multiplyby (1 + t)γk , and make the sum over k to obtain:

dZ

dt(t) +

a

(1 + t)Z(t) ≤ C

(1

(1 + t)2Z(t) +

∑k

∑x∈Ek

Z(t)2+x(1 + t)β+ax

).

1430 M. Grassin

The constant C is positive and depends only on γ, m, d, δ, ‖u0‖X . We used that

β = −γ1 −d

2= −1− r + a−

d

2= −1−

b

2+ a.

Now we claim that

dZ

dt(t) +

a

(1 + t)Z(t) ≤ C

(1

(1 + t)2Z(t) + Z(t)2(1 + t)β

+ Z(t)2+m(1 + t)β+am

),

since we keep the smaller and the biggest e(x) in the sum. We choose a = 1+b/2,that is to say β = 0 to obtain

dZ

dt(t) +

a

(1 + t)Z(t) ≤ C

(1

(1 + t)2Z(t) + Z(t)2(27)

+ Z(t)2+m(1 + t)am).

We set ζ(t) = (1 + t)a exp(C/(1 + t))Z(t) to get

dζ

dt(t) ≤ C

ζ2

(1 + t)a(1 + ζ(t)m).(28)

Using that ζ2 + ζ2+m ≤ 2ζ(1 + ζ1+m), we modify (28)

1

ζ(1 + ζ1+m)

dζ

dt(t) ≤

C

(1 + t)a.(29)

Now we can integrate (29). Consider

f(x) =1

m+ 1ln

(xm+1

(1 + xm+1)

)and ψ(t) = f(ζ(t)) +

C

a− 1(1 + t)1−a.

We derivate ψ to obtain

ψ′(t) =1

ζ(1 + ζ1+m)ζ ′(t)−

C

(1 + t)a.

Since (29) holds, we conclude that:

ψ(t) ≤ ψ(0) for all t ≥ 0.

And this leads to

f(ζ(t)) ≤ f(ζ(0)) +C

a− 1for all t ≥ 0.


To finish, we remark that f is strictly increasing and that f(x)→ −∞ whenx→ 0. Thus if ζ(0) ≤ ε0, then

f(ζ(t)) ≤ f(ε0) +C

a− 1for all t ≥ 0.

This implies that ζ(t) ≤M0 = M(ε0, a, C) for all t ≥ 0. As a consequence, wehave:

Z(t) ≤M0(1 + t)−a exp

(−

C

1 + t

)for all t ≥ 0.

We conclude that if we have Z(0) ≤ ε0 exp(−C), then

Z(t) ≤M0(1 + t)−a for all t ≥ 0.

That corresponds to hypothesis (H1), since Z(0) ∼ ‖π0‖m + ‖S0‖m.

2.2. Uniqueness results and corollary. In the isentropic case, we provetwo results of uniqueness: the first one is local in time and space, the second oneis local in space and global in time. In the proofs, we used the symmetric formof the system (21):

A0(V )∂tV +d∑

α=1

Aα(V )∂αV = 0.(30)

The proofs work in the same way in the general case, one has just to deal withA0(V ) instead of A0.

Proposition 7. Let V 10 , V 2

0 be two initial data for (21). Assume thatV 1

0 ∈ Hm(Rd). Let V 1 = (π, u, S)T , V 2 be two associated solutions of (21)defined for 0 ≤ t ≤ T0 and let

M ≥ sup0≤t≤T0

(C1e

|S0|∞/2γ |π|∞ + |u|∞)(t).

Suppose that V 10 = V 2

0 on B0 = B(x0, η). We take CT = {(x, t) | 0 ≤ t ≤ T ,x ∈ Bt = B(x0, η −Mt)} for 0 ≤ T ≤ T1 = min(T0, η/M). Then V 1 = V 2 onCT1

.

Proposition 8. Let (ρ0, u0, S0) satisfy (H1)-(H4). Let U = (π, u, S)T bethe global solution of (21) given by Theorem (3) and u be the solution of (4).Consider V a global solution of (21) such that DV ∈ L∞(Rd × R+).

Then for all ν ∈ ]2−a, 1[, for all R0 > 0, there exists T0 > 0 such that, ifU(·, T0) = V (·, T0) on B(0, R0), then U and V are equal on the domain {(x, t) :|x− x(t)| ≤ R(t), for all t ≥ T0}, where x(t) is the solution of x′(t) = u(x(t), t),x(T0) = 0, and R(t) = R0(1 + t)ν .

Using the local uniqueness result, we can show as in the isentropic case thatthe hypothesis (H4) can be forgotten:

Corollary 2. Let (ρ0, u0, S0) be the initial data for Problem (1). Suppose(H1)-(H3). Then the result of Theorem 3 is still true.

1432 M. Grassin

References

[1] D. Serre, Solutions classiques globales des equations d’Euler pour unfluide parfait compressible, Annales de l’Institut Fourier 47 (1952), 139–153.

[2] J. Y. Chemin, Dynamique des gaz a masse totale finie, Asymptotic Anal-ysis 3 (1990), 215–220.

[3] P. D. Lax, Development of singularities of solutions of nonlinear hyper-bolic partial differential equations, J. Math. Phys. 5 (1964), 611–513.

[4] T. Sideris, Formation of singularities in three-dimensional compressiblefluids, Commun. Math. Phys. 101 (1985), 475–485.

[5] Liu Fagui, Global smooth resolvability for one dimensional gas dynamicssystems, communication personnelle.

[6] T. Makino, S. Ukai & S. Kawashima, Sur la solution a support com-pact de l’equation d’Euler compressible, Japan J. Appl. Math. 3 (1986),249–257.

[7] A. Friedman, Partial Differential Equations, Holt, Rinehart, Winston,1969.

UMPACNRS–ENS Lyon69364 Lyon cedex 07FRANCE

Email: [email protected]

Received: April 10th, 1998; revised: June 22nd, 1998.

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